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I have found an example: What is $\displaystyle\lim_{n\rightarrow\infty}\int_{0}^{\infty}\frac{1}{n}\chi_{[0,n)}\,dx$ ?

I mean, I know it is $1$, because for every $n$, the value of the integral is $1$, and the limit of the constant $1$ function is also $1$. However, I don't understand why I can't use the Dominated Convergence Theorem!? If I change the $\lim$ and $\int$ operators, I get $0$, since $\lim_{n\rightarrow\infty}\frac{1}{n}\chi_{[0,n)}=0$, and if I integrate $0$, it is going to be $0$.

Therefore, I miss something from the Dominated Convergence Theorem, or I don't understand it clearly. Why can't I change the the order of operators?

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  • $\begingroup$ You need a dominating function which is integrable. $\endgroup$ – Thomas Shelby Aug 7 '19 at 8:47
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DCT requires the condition that the integrand is dominated by an integarble function. In this case if $\frac 1 n I_{[n, \infty)} \leq g$ then $g \geq \frac 1 n$ on the interval $[n,\infty)$ for all $n$ and, in particular for $n=1$, so $\int_0^{\infty} g =\infty$. Thus there is no dominating integrable function,.

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